# Alternate definition of higher characteristic classes

While I was trying to find out why the first stiefel whitney class of a vector bundle is zero iff its orientable, I figured out that $w_1(Bundle)=w_1(Top\,exterior\,power\,of\,bundle)$ which is zero if the vector bundle is orientable. I did this using the kunneth formula for the exterior power of a direct sum and the formula for the characteristic classes of a tensor product of line bundles.

Question: Given a real vector bundle $E \to B$, is $w_i(E)=w_1(\Lambda^i E)$?

• The equation you've written doesn't make sense, since $w^i\in H^i$ and $w_1\in H^1$. – Eric Wofsey Jun 23 '17 at 19:54
• sorry :) can I delete this question? – user062295 Jun 23 '17 at 19:55
• Given that $w_1(E) = w_1(\bigwedge^{\operatorname{rank}E}E)$, it seems that the natural question to ask is whether $w_1(E) = w_1(\bigwedge^i E)$. Is that what you meant to ask? – Michael Albanese Jun 24 '17 at 5:02